Solve for $x$ : $ 6|x + 2| - 5 = 4|x + 2| + 6 $
Subtract $ {4|x + 2|} $ from both sides: $ \begin{eqnarray} 6|x + 2| - 5 &=& 4|x + 2| + 6 \\ \\ { - 4|x + 2|} && { - 4|x + 2|} \\ \\ 2|x + 2| - 5 &=& 6 \end{eqnarray} $ Add ${5}$ to both sides: $ \begin{eqnarray} 2|x + 2| - 5 &=& 6 \\ \\ { + 5} &=& { + 5} \\ \\ 2|x + 2| &=& 11 \end{eqnarray} $ Divide both sides by ${2}$ $ \dfrac{2|x + 2|} {{2}} = \dfrac{11} {{2}} $ Simplify: $ |x + 2| = \dfrac{11}{2}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 2 = -\dfrac{11}{2} $ or $ x + 2 = \dfrac{11}{2} $ Solve for the solution where $x + 2$ is negative: $ x + 2 = -\dfrac{11}{2} $ Subtract ${2}$ from both sides: $ \begin{eqnarray} x + 2 &=& -\dfrac{11}{2} \\ \\ {- 2} && {- 2} \\ \\ x &=& -\dfrac{11}{2} - 2 \end{eqnarray} $ Change the ${ - 2}$ to an equivalent fraction with a denominator of $2$ $ x = - \dfrac{11}{2} {- \dfrac{4}{2}} $ $ x = -\dfrac{15}{2} $ Then calculate the solution where $x + 2$ is positive: $ x + 2 = \dfrac{11}{2} $ Subtract ${2}$ from both sides: $ \begin{eqnarray} x + 2 &=& \dfrac{11}{2} \\ \\ {- 2} && {- 2} \\ \\ x &=& \dfrac{11}{2} - 2 \end{eqnarray} $ Change the ${ - 2}$ to an equivalent fraction with a denominator of $2$ $ x = \dfrac{11}{2} {- \dfrac{4}{2}} $ $ x = \dfrac{7}{2} $ Thus, the correct answer is $x = -\dfrac{15}{2} $ or $x = \dfrac{7}{2} $.